cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A239036 A set of eleven distinct positive odd numbers the sum of whose reciprocals is 1 and whose 11th term is as large as possible.

Original entry on oeis.org

3, 5, 7, 9, 11, 13, 23, 721, 979011, 175878510309, 20622166925499467673345
Offset: 1

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Author

Arkadiusz Wesolowski, Mar 09 2014

Keywords

Comments

If k is the largest number in the set of eleven distinct positive odd numbers the sum of whose reciprocals is 1, then k <= a(11).
Is there any set of eleven distinct positive odd numbers the sum of whose reciprocals is 1 and having the Egyptian number greater than 20622166925675347163457?
This is similar to the problem discussed by Curtiss (see link), but the numbers are restricted to be odd. - T. D. Noe, Mar 18 2014

Examples

			1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/23 + 1/721 + 1/979011 + 1/175878510309 + 1/20622166925499467673345 = 1.
		

Crossrefs

Programs

  • PARI
    f=0; n=3; s=11; if(s<11, break); for(t=1, s-3, print1(n, ", "); f=f+1/n; until(1>f+1/n, n=n+2)); until(numerator(1-f-1/n)==2, n=n+2); print1(n, ", "); f=f+1/n; g=2*floor((numerator(f)+1)/4)+1; until(numerator(1-f-1/g)==1, g=g+2); print1(g, ", "); f=f+1/g; print1(denominator(1-f));

A351532 Number of integer pairs (i, j), 0 < i, j < n, such that i/(n - i) + j/(n - j) = 1.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 2, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 2, 5, 0, 0, 1, 2, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 2, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 5, 0, 0, 1, 0, 2, 1, 0, 2, 1, 0, 0, 3, 0, 0, 1, 0, 0, 7, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 1, 0, 2, 3
Offset: 1

Views

Author

Lars Blomberg, Feb 14 2022

Keywords

Comments

By symmetry, if (i, j) is a solution then so is (j, i). When j=i we get n = 3i, corresponding to the solution 1/2 + 1/2 = 1. Therefore, when 3|n, a(n) > 0 and odd, otherwise a(n) >= 0 and even.
For n < 10^6, the largest term is a(720720) = 285, and 483188 terms are 0.
Other record terms: a(1081080) = 369, a(2162160) = 457, a(3243240) = 481, a(4324320) = 533, a(5405400) = 559, a(6126120) = 597. Record terms with index >= 360360 appear to occur at indices that are multiples of 180180. - Chai Wah Wu, Feb 15 2022

Examples

			For n = 3: (i, j) = (1, 1), so a(3) = 1. (1/2 + 1/2 = 1)
For n = 18: (i, j) = (3, 8), (6, 6), (8, 3), so a(18) = 3. (3/15 + 8/10 = 1/5 + 4/5 = 1)
For n = 20: (i, j) = (5, 8), (8, 5), so a(20) = 2.
For n = 36: (i, j) = (6, 16), (8, 15), (12, 12), (15, 8), (16, 6), so a(36) = 5.
		

Crossrefs

Programs

  • PARI
    a(n)={my(x=n^2, y=2*n); sum(i=1,(n-1)/2, x-=2*n; y-=3; if(x%y==0,1,0))}
    
  • Python
    from sympy.abc import i, j
    from sympy.solvers.diophantine.diophantine import diop_quadratic
    def A351532(n):
        return sum(1 for d in diop_quadratic(n**2+3*i*j-2*n*(i+j)) if 0 < d[0] < n and 0 < d[1] < n) # Chai Wah Wu, Feb 15 2022

Formula

The original equation can be solved for j giving j = (n(n - 2i)) / (2n - 3i). Varying i from 1 to n-1, a(n) is given by the number of integer values of j, 0 < j < n.

Extensions

Data section extended up to a(105) by Antti Karttunen, Jan 17 2025

A240132 Number of solutions of 1/x_1 + 1/x_2 + ... + 1/x_{2n-1} = 1 in distinct odd numbers > 1.

Original entry on oeis.org

0, 0, 0, 0, 5, 379118
Offset: 1

Views

Author

Jonathan Sondow, Apr 21 2014

Keywords

Comments

The next term is a(7) > 467871307.

Examples

			If 1 < x_1 < x_2 < x_3 are odd, then 1/x_1 + 1/x_2 + 1/x_3 < 3/3 = 1, so a(2) = 0.
		

Crossrefs

The five sets of 9 odd numbers are: A201644, A201646, A201647, A201648, A201649.
See A238795 and A239036 for specific sets of 11 odd numbers.
Showing 1-3 of 3 results.